In triangle CDE EC = 26 cm, segment MN is parallel to CE, M belongs to CD, N belongs to ED. Find CD if CM = 8 cm, MN = 20 cm.
Let us prove that the triangles CDE and MDN are similar.
In triangles, the angle at the vertex D is common, and the angles DCE and DMN are equal, as are the corresponding angles, at the intersection of the secant CD parallel to CE and MN. Then the triangles CDE and MDN are similar in the first sign of similarity.
Let the segment DМ = X cm, then СD = (X + 8).
EC / NM = DC / DM.
26/20 = (X + 8) / X.
20 * X + 160 = 26 * X.
6 * X = 160.
X = 160/6 = 80/3.
Then SD = 80/3 + 8 = 104/3 = 34 (2/3) cm.
Answer: SD = 34 (2/3) cm.
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